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3 years ago

Please help, will mark brainliest!!

Mathematics

1 answer:

Novay_Z [31]3 years ago

3 0

Answer:

$12000

Step-by-step explanation:

P=CRT

C=cost

R=rate

T=time

P=3000 x 8 x 50/100= $12000

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Jane had 275 stickers, which she gave to her friends. Each friend got five more stickers than the friend before. If the first fr

Paha777 [63]

Answer:

10 friends

Step-by-step explanation:

we know that

The formula of the sum is equal to

$sum=\frac{n}{2}[2a1+(n-1)d]$

where

a1 is the first term

n is the number of terms (number of friends)

d is the common difference in the arithmetic sequence

In this problem we have

$sum=275\ stickers$

$a1=5\ stickers$

$d=5$ ----> the common difference

substitute in the formula and solve for n

$275=\frac{n}{2}[2(5)+(n-1)(5)]$

$550=n[10+5n-5]\\ \\550=10n+5n^{2} -5n\\ \\5n^{2}+5n-550=0$

Solve the quadratic equation by graphing

The solution is n=10

see the attached figure

therefore

She had 10 friends who got stickers

5 0

3 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

4 years ago

Let E1 be the event that tails come up when the coin is tossed the first time and E2 be the event that heads come up when the co

Norma-Jean [14]

Answer:

Step-by-step explanation:

In the tossing of a fair coin, there are equal probabilities of getting a HEAD and getting a TAIL.

Total probability is always 1 and a coin has 2 faces - Head & Tail.

The probability of getting a Head is 1/2 = 0.5

The probability of getting a Tail is 1/2 = 0.5

E1 is the event that TAIL comes up when the coin is tossed the first time

E2 is the event that HEAD comes up when the coin is tossed the second time

The probability value for EVENT 1 is 0.5

The probability value for EVENT 2 is 0.5

8 0

4 years ago

Write an expression for the product of - 3 *- 2y

Wewaii [24]

I think it’s 6y hopefully it’s correct lol

3 0

3 years ago

Find the area of this trapezoid. no links only answer

-Dominant- [34]

Answer:

221

Step-by-step explanation: 5x10=50 9x19=171 171+50=221

5 0

3 years ago